We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. SQL
  3. Advanced Select
  4. The PADS
  5. Discussions

The PADS

Sort by

recency

|

5593 Discussions

|

  • luckysinghsolut1
     + 0 comments
    SET NOCOUNT ON;

(SELECT CONCAT(NAME, "(", LEFT(OCCUPATION,1), ")") AS OC
FROM OCCUPATIONS
)

UNION 

(SELECT CONCAT("There are a total of ", COUNT(OCCUPATION) OVER(PARTITION BY OCCUPATION)," ", LOWER(OCCUPATION),"s.") AS OC 
FROM OCCUPATIONS
); 

go

  • nandakumar_usao1
     + 0 comments

    EASIEST WAY OF WRITING THIS QUERY DYNAMICALLY!!

    SELECT NAME || '(' || SUBSTR(OCCUPATION, 1, 1) || ')' FROM OCCUPATIONS ORDER BY NAME;

    SELECT
    'There are a total of ' || COUNT(STAR) || ' ' || LOWER(OCCUPATION) || 's.' FROM OCCUPATIONS GROUP BY OCCUPATION ORDER BY COUNT(STAR) ASC, OCCUPATION ASC;

  • abhishekkapoor81
     + 0 comments

    EASIEST Query:

    SELECT 
    CONCAT(NAME, '(', SUBSTR(OCCUPATION, 1, 1), ')') AS RESULT
FROM 
    OCCUPATIONS ORDER BY NAME;


SELECT  
    CONCAT('There are a total of ', COUNT(*), ' ', LOWER(OCCUPATION), 's.') AS RESULT
FROM 
    OCCUPATIONS 
GROUP BY 
    OCCUPATION 
ORDER BY 
    COUNT(OCCUPATION), OCCUPATION;

  • mo_nagah2003
     + 0 comments

    SELECT CONCAT(name, '(', LEFT(occupation,1), ')') FROM occupations ORDER BY name SELECT CONCAT('There are a total of ', COUNT(occupation), ' ', LOWER(occupation), 's.') FROM occupations GROUP BY occupation ORDER BY COUNT(occupation), occupation

  • rjd19972
     + 0 comments

    SELECT concat(name, '(',left(occupation, 1),')') from occupations order by name; select concat('There are a total of ', count(),' ', lower(occupation),'s.') from occupations group by occupation order by count() asc, occupation asc;